Question

If all the three altitudes of a triangle are equal, then prove that it is a isosceles triangle no spamming pls

Answer 1

Answer:

Consider triangles BEC and BFC

EC=BF (Equal altitudes: Given)

∠BEC=∠BFC=90 (BE and BF are Altitudes)

BC = BC (common)

△BEC≅△BFC (RHS postulate)

∠ABC=∠BCA (Corresponding angles)

Consider the △ ADB and △ ADC

∠ADB=∠ADC=90 (AD is Altitude)

AD = AD (common)

∠ABC=∠BCA

Thus, △ADB≅△ADC (ASA postulate)

AB =AC (Corresponding sides are equal)

BC= AC (Similarly, we can prove △BFC≅△BFA)

Thus, AB=AC=BC

Step-by-step explanation:

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Answer 2

To prove:

If all the 3 altitudes of a Triangle are equal , then the Triangle is equilateral ∆.

Proof:

Let's consider a ∆ABC , such that AD , BF and CE are the altitudes and they are equal ;

In ∆BEC and ∆BCF:

1) $\angle$BEC = $\angle$CFB=90°

2) CE = BF (given)

3) BC is common side.

Hence , ∆BEC $\cong$ ∆BCF (RHS congruency)

Therefore $\angle ABC = \angle BCA$ (as per CPCT)

In ∆ABD and ∆ACD:

1) AD is common side

2) $\angle ADB = \angle ADC$=90°

3) $\angle ABC = \angle BCA$ (proved earlier)

Hence ∆ABD $\cong$ACD (ASA congruency)

So, AB = AC (As per CPCT)

Now , we can also prove BC = AC ,

( because ∆BCF $\cong$ ∆BAF)

So, AB = AC = BC

Hence Triangle is Equilateral.